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How many elements there exist in quotient ring $\mathbb{Z}_5[X]/(X^2+1)$?

I'm learning polynomial ring. However I cant' completely understand the number of elements. I think that the elements are contained as follows. \begin{align} \mathbb{Z}_5[X]/(X^2+1)=\{&0,1,2,3,4,x,x+1,x+2,x+3,x+4,\\ & 2x,2x+1,2x+2,2x+3,2x+4,3x,3x+1,3x+2,3x+3,3x+4,\\ & 4x,4x+1,4x+2,4x+3,4x+4,x^2 \} \end{align}

There are 26 elements in total, is it correct?

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    $x^2=-1=4$ so you've counted that one twice. – 2017-02-17
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    thanks @ziggurism :) so 25 elements do in total ♡ – 2017-02-17
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    Yes, 25 is correct. It's a vector space of dimension 2 over $\mathbb{Z}/5$. In general a vector space of dimension $d$ over a field $k$ has cardinality $\lvert k\rvert^d$. – 2017-02-17

2 Answers 2

5

You are correct but there is a less tedious way of doing this. You can notice that any polynomial $f(x) = \sum a_n x^n$ can be divided by $x^2 + 1$, i.e be written $f(x) = (x^2 + 1)q(x) + r(x)$ with $\deg(r) < 2$.

By definition $f(x) = r(x)$ in the quotient ring, and it its clear that two differents linear polynomials (with coefficient modulo 5) are differents. So the number of elements in the quotient rings is the number of linear polynomial with coefficient in $\mathbb F_5$ which is $5 \times 5 = 25$.

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Since $X^2+1=(X-2)(X-3)$ with $\gcd(X-2,X-3)=1$, we have $$\mathbb{Z}_5[X]/\left(X^2+1\right)\cong \Big(\mathbb{Z}_5[X]/(X-2)\Big)\oplus \Big(\mathbb{Z}_5[X]/(X-3)\Big)\cong \mathbb{Z}_5\oplus\mathbb{Z}_5\,,$$ as rings. In fact, one does not need to factorize $X^2+1$. If $f(X)\in\mathbb{K}[X]$ is of degree $d$, where $\mathbb{K}$ is a field, then $$\dim_\mathbb{K}\Big(\mathbb{K}[X]/\big(f(X)\big)\Big)=d\,.$$